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According to the second postulate of Special Relativity the speed of light c must be invariant that consequently leads to the first postulate of Einstein the Lorentz invariance in all inertial frames of reference.

However, with ZPF energy fluctuations of vacuum space with virtual particles popping out of the vacuum from 'nothing', I find it difficult to define ZPF as Lorentz invariant.

Unless it is conclusively proven that the non-zero tiny ZPF noise energy is due to almost perfect QFT fields cancellation it is difficult to claim that ZPF of vacuum is Lorentz invariant.

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  • $\begingroup$ The issue is discussed here: physics.stackexchange.com/questions/447371/… . $\endgroup$
    – Andrea Alciato
    Feb 19 at 8:52
  • $\begingroup$ @AndreaAlciato This question uses a different context from the question you have referenced. Namely, it asks in order prove that ZPF is Lorentz Invartiant first you have to prove that it is a product of QFT's quantum fields almost completelly cancelling out in free space with a tiny residue left out thus ZPF noise. I don't know if such a proof is possible and also backed up by any empirical data. Alternativelly it could be that ZPF of the vacuum is dark energy noise in our spacetime. Since we have not a complete theory about dark energy then we cannot even claim that ZPF is Lorentz Invariant. $\endgroup$
    – Markoul11
    Feb 19 at 18:00
  • $\begingroup$ This paper here speaks clearly about the ZPF zero point energy fluctuations of the vacuum exhibiting stochastic Lorentz- Invariance violation: nature.com/articles/nphys3270 $\endgroup$
    – Markoul11
    Mar 20 at 6:05

2 Answers 2

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In technical terms, the zero-point-energy is the counterterm of the form $\mathcal L_\text{ct}\supset\text{const}$. Given that the Lagrangian is a Lorentz scalar by definition, this makes it manifest that this counterterm is a scalar as well.

Being a counterterm, it is not scheme-independent and thus not measurable by itself, although for some suitable backgrounds it does have an observable effect. Flat spacetime is too boring of a background and does not allow you to detect any physical effects of this couterterm. That being said, in flat spacetime the contribution of this counterterm to the momentum operator is $$ P_\mu\sim \int \delta(p^2-m^2)p_\mu\mathrm dp $$ For $\mu=i$ this integral vanishes by spherical symmetry, and therefore only the $\mu=0$ part is non-trivial. This is the reason we call this a zero-point-energy. Lorentz invariance is perhaps non-manifest since only the $\mu=0$ part is non-zero. But of course in more interesting backgrounds, there is also a non-trivial contribution to the momentum (which is essential, for example, in $2d$ CFTs), and Lorentz invariance is more manifest. In any case, Lorentz invariance is always guaranteed to hold given the definition in terms of $\mathcal L_\text{ct}$.

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  • $\begingroup$ In order prove that ZPF is Lorentz Invartiant first you have to prove that it is a product of QFT's quantum fields almost completelly cancelling out in free space with a tiny residue left out thus ZPF noise. I don't know is such a proof is possible and also backed up by any empirical data. On the other hand it could be simply that ZPF of the vacuum is dark energy noise in our spacetime. If this is the case since we have not a complete theory about dark energy we cannot even claim that ZPF is Lorentz Invariant. $\endgroup$
    – Markoul11
    Feb 19 at 17:52
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Both postulates are independent claims, it is not true that the second leads to the first. This is easy to see when we apply the same logic to wave propagation in ether-like systems, e.g. sound propagation in atmosphere. Sound speed is independent of the relative motion of the source, but this does not imply that all frames in the atmosphere are equivalent.

Zero point EM radiation is usually described by a spectral function of frequency (Poynting energy density per unit frequency interval) that is proportional to 3rd power of frequency. People who studied the concept of zero point radiation in depth (e.g. Timothy Boyer) claim this spectral function is Lorentz-invariant [1]. Virtual particles are artifacts of perturbation formalism, they do not "pop out of the vacuum" in reality.

[1] T. H. Boyer, Random electrodynamics: The theory of classical electrodynamics with classical electromagnetic zero-point radiation, Phys. Rev. D 11, 790-808 (1975). (Lorentz invariance is discussed in Section D.)

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  • $\begingroup$ Reality is tricky. "Squeezed vacuum" is a real thing, with observed consequences, created by controlling zero point fluctuations. $\endgroup$
    – John Doty
    Feb 9 at 14:57
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    $\begingroup$ @JohnDoty Not sure I understand your point. Of course squeezed states of light can be prepared. Whether the resulting EM field is called "squeezed vacuum" or superposition of zero point radiation and emission due to material sources (laser, crystal) does not seem relevant here. On the fundamental level, zero point radiation (in theories that assume its existence) cannot be controlled, because it is that part of the field that is omnipresent no matter what we do, it is a "boundary condition". $\endgroup$
    – Ján Lalinský
    Feb 9 at 15:11
  • $\begingroup$ Related to "Virtual particles are artifacts". They are just as artificial as (and related to) squeezed vacuum. But all mathematical objects used to capture physical reality are artificial: as mathematics, they exist only in the human mind. $\endgroup$
    – John Doty
    Feb 9 at 15:19
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    $\begingroup$ No, some mathematical descriptions are more fundamental than others. EM field is fundamental entity that can be measured, virtual particles are metaphors that describe terms in calculations. $\endgroup$
    – Ján Lalinský
    Feb 9 at 15:24
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    $\begingroup$ Casimir force alone is not strong evidence for virtual particles, or even zero point radiation. It can be explained alternatively as result of EM interaction of atoms in the walls (van der Waals forces). Don't rely on wikipedia for learning physics, there is lots of errors and misconceptions there, prefer books and papers. $\endgroup$
    – Ján Lalinský
    Feb 10 at 16:43

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